package geogebra.kernel.jama;
import geogebra.kernel.jama.util.Maths;

/**
 * Singular Value Decomposition.
 * <P>
 * For an m-by-n matrix A with m >= n, the singular value decomposition is an
 * m-by-n orthogonal matrix U, an n-by-n diagonal matrix S, and an n-by-n
 * orthogonal matrix V so that A = U*S*V'.
 * <P>
 * The singular values, sigma[k] = S[k][k], are ordered so that sigma[0] >=
 * sigma[1] >= ... >= sigma[n-1].
 * <P>
 * The singular value decompostion always exists, so the constructor will never
 * fail. The matrix condition number and the effective numerical rank can be
 * computed from this decomposition.
 */

class SingularValueDecomposition implements java.io.Serializable {

  /*
   * ------------------------ Class variables ------------------------
   */

  /**
     * 
     */
  private static final long serialVersionUID = -6468895344826888048L;

  /**
   * Arrays for internal storage of U and V.
   * 
   * @serial internal storage of U.
   * @serial internal storage of V.
   */
  private final double[][] U, V;

  /**
   * Array for internal storage of singular values.
   * 
   * @serial internal storage of singular values.
   */
  private final double[] s;

  /**
   * Row and column dimensions.
   * 
   * @serial row dimension.
   * @serial column dimension.
   */
  private final int m, n;

  /*
   * ------------------------ Constructor ------------------------
   */

  /**
   * Construct the singular value decomposition
   * 
   * @param A
   *          Rectangular matrix
   * @return Structure to access U, S and V.
   */

  public SingularValueDecomposition(Matrix Arg) {

    // Derived from LINPACK code.
    // Initialize.
    double[][] A = Arg.getArrayCopy();
    m = Arg.getRowDimension();
    n = Arg.getColumnDimension();

    /*
     * Apparently the failing cases are only a proper subset of (m<n), so let's
     * not throw error. Correct fix to come later? if (m<n) { throw new
     * IllegalArgumentException("Jama SVD only works for m >= n"); }
     */
    int nu = Math.min(m, n);
    s = new double[Math.min(m + 1, n)];
    U = new double[m][nu];
    V = new double[n][n];
    double[] e = new double[n];
    double[] work = new double[m];
    boolean wantu = true;
    boolean wantv = true;

    // Reduce A to bidiagonal form, storing the diagonal elements
    // in s and the super-diagonal elements in e.

    int nct = Math.min(m - 1, n);
    int nrt = Math.max(0, Math.min(n - 2, m));
    for (int k = 0; k < Math.max(nct, nrt); k++) {
      if (k < nct) {

        // Compute the transformation for the k-th column and
        // place the k-th diagonal in s[k].
        // Compute 2-norm of k-th column without under/overflow.
        s[k] = 0;
        for (int i = k; i < m; i++)
          s[k] = Maths.hypot(s[k], A[i][k]);
        if (s[k] != 0.0) {
          if (A[k][k] < 0.0)
            s[k] = -s[k];
          for (int i = k; i < m; i++)
            A[i][k] /= s[k];
          A[k][k] += 1.0;
        }
        s[k] = -s[k];
      }
      for (int j = k + 1; j < n; j++) {
        if (k < nct & s[k] != 0.0) {

          // Apply the transformation.

          double t = 0;
          for (int i = k; i < m; i++)
            t += A[i][k] * A[i][j];
          t = -t / A[k][k];
          for (int i = k; i < m; i++)
            A[i][j] += t * A[i][k];
        }

        // Place the k-th row of A into e for the
        // subsequent calculation of the row transformation.

        e[j] = A[k][j];
      }
      if (wantu & k < nct)
        for (int i = k; i < m; i++)
          U[i][k] = A[i][k];
      if (k < nrt) {

        // Compute the k-th row transformation and place the
        // k-th super-diagonal in e[k].
        // Compute 2-norm without under/overflow.
        e[k] = 0;
        for (int i = k + 1; i < n; i++)
          e[k] = Maths.hypot(e[k], e[i]);
        if (e[k] != 0.0) {
          if (e[k + 1] < 0.0)
            e[k] = -e[k];
          for (int i = k + 1; i < n; i++)
            e[i] /= e[k];
          e[k + 1] += 1.0;
        }
        e[k] = -e[k];
        if (k + 1 < m & e[k] != 0.0) {

          // Apply the transformation.

          for (int i = k + 1; i < m; i++)
            work[i] = 0.0;
          for (int j = k + 1; j < n; j++)
            for (int i = k + 1; i < m; i++)
              work[i] += e[j] * A[i][j];
          for (int j = k + 1; j < n; j++) {
            double t = -e[j] / e[k + 1];
            for (int i = k + 1; i < m; i++)
              A[i][j] += t * work[i];
          }
        }
        if (wantv)
          for (int i = k + 1; i < n; i++)
            V[i][k] = e[i];
      }
    }

    // Set up the final bidiagonal matrix or order p.

    int p = Math.min(n, m + 1);
    if (nct < n)
      s[nct] = A[nct][nct];
    if (m < p)
      s[p - 1] = 0.0;
    if (nrt + 1 < p)
      e[nrt] = A[nrt][p - 1];
    e[p - 1] = 0.0;

    // If required, generate U.

    if (wantu) {
      for (int j = nct; j < nu; j++) {
        for (int i = 0; i < m; i++)
          U[i][j] = 0.0;
        U[j][j] = 1.0;
      }
      for (int k = nct - 1; k >= 0; k--)
        if (s[k] != 0.0) {
          for (int j = k + 1; j < nu; j++) {
            double t = 0;
            for (int i = k; i < m; i++)
              t += U[i][k] * U[i][j];
            t = -t / U[k][k];
            for (int i = k; i < m; i++)
              U[i][j] += t * U[i][k];
          }
          for (int i = k; i < m; i++)
            U[i][k] = -U[i][k];
          U[k][k] = 1.0 + U[k][k];
          for (int i = 0; i < k - 1; i++)
            U[i][k] = 0.0;
        } else {
          for (int i = 0; i < m; i++)
            U[i][k] = 0.0;
          U[k][k] = 1.0;
        }
    }

    // If required, generate V.

    if (wantv)
      for (int k = n - 1; k >= 0; k--) {
        if (k < nrt & e[k] != 0.0)
          for (int j = k + 1; j < nu; j++) {
            double t = 0;
            for (int i = k + 1; i < n; i++)
              t += V[i][k] * V[i][j];
            t = -t / V[k + 1][k];
            for (int i = k + 1; i < n; i++)
              V[i][j] += t * V[i][k];
          }
        for (int i = 0; i < n; i++)
          V[i][k] = 0.0;
        V[k][k] = 1.0;
      }

    // Main iteration loop for the singular values.

    int pp = p - 1;
    int iter = 0;
    double eps = Math.pow(2.0, -52.0);
    double tiny = Math.pow(2.0, -966.0);
    while (p > 0) {
      int k, kase;

      // Here is where a test for too many iterations would go.

      // This section of the program inspects for
      // negligible elements in the s and e arrays. On
      // completion the variables kase and k are set as follows.

      // kase = 1 if s(p) and e[k-1] are negligible and k<p
      // kase = 2 if s(k) is negligible and k<p
      // kase = 3 if e[k-1] is negligible, k<p, and
      // s(k), ..., s(p) are not negligible (qr step).
      // kase = 4 if e(p-1) is negligible (convergence).

      for (k = p - 2; k >= -1; k--) {
        if (k == -1)
          break;
        if (Math.abs(e[k]) <= tiny + eps
            * (Math.abs(s[k]) + Math.abs(s[k + 1]))) {
          e[k] = 0.0;
          break;
        }
      }
      if (k == p - 2)
        kase = 4;
      else {
        int ks;
        for (ks = p - 1; ks >= k; ks--) {
          if (ks == k)
            break;
          double t = (ks != p ? Math.abs(e[ks]) : 0.)
              + (ks != k + 1 ? Math.abs(e[ks - 1]) : 0.);
          if (Math.abs(s[ks]) <= tiny + eps * t) {
            s[ks] = 0.0;
            break;
          }
        }
        if (ks == k)
          kase = 3;
        else if (ks == p - 1)
          kase = 1;
        else {
          kase = 2;
          k = ks;
        }
      }
      k++;

      // Perform the task indicated by kase.

      switch (kase) {

        // Deflate negligible s(p).

        case 1 : {
          double f = e[p - 2];
          e[p - 2] = 0.0;
          for (int j = p - 2; j >= k; j--) {
            double t = Maths.hypot(s[j], f);
            double cs = s[j] / t;
            double sn = f / t;
            s[j] = t;
            if (j != k) {
              f = -sn * e[j - 1];
              e[j - 1] = cs * e[j - 1];
            }
            if (wantv)
              for (int i = 0; i < n; i++) {
                t = cs * V[i][j] + sn * V[i][p - 1];
                V[i][p - 1] = -sn * V[i][j] + cs * V[i][p - 1];
                V[i][j] = t;
              }
          }
        }
          break;

        // Split at negligible s(k).

        case 2 : {
          double f = e[k - 1];
          e[k - 1] = 0.0;
          for (int j = k; j < p; j++) {
            double t = Maths.hypot(s[j], f);
            double cs = s[j] / t;
            double sn = f / t;
            s[j] = t;
            f = -sn * e[j];
            e[j] = cs * e[j];
            if (wantu)
              for (int i = 0; i < m; i++) {
                t = cs * U[i][j] + sn * U[i][k - 1];
                U[i][k - 1] = -sn * U[i][j] + cs * U[i][k - 1];
                U[i][j] = t;
              }
          }
        }
          break;

        // Perform one qr step.

        case 3 : {

          // Calculate the shift.

          double scale = Math.max(Math.max(Math.max(Math.max(
              Math.abs(s[p - 1]), Math.abs(s[p - 2])), Math.abs(e[p - 2])),
              Math.abs(s[k])), Math.abs(e[k]));
          double sp = s[p - 1] / scale;
          double spm1 = s[p - 2] / scale;
          double epm1 = e[p - 2] / scale;
          double sk = s[k] / scale;
          double ek = e[k] / scale;
          double b = ((spm1 + sp) * (spm1 - sp) + epm1 * epm1) / 2.0;
          double c = sp * epm1 * sp * epm1;
          double shift = 0.0;
          if (b != 0.0 | c != 0.0) {
            shift = Math.sqrt(b * b + c);
            if (b < 0.0)
              shift = -shift;
            shift = c / (b + shift);
          }
          double f = (sk + sp) * (sk - sp) + shift;
          double g = sk * ek;

          // Chase zeros.

          for (int j = k; j < p - 1; j++) {
            double t = Maths.hypot(f, g);
            double cs = f / t;
            double sn = g / t;
            if (j != k)
              e[j - 1] = t;
            f = cs * s[j] + sn * e[j];
            e[j] = cs * e[j] - sn * s[j];
            g = sn * s[j + 1];
            s[j + 1] = cs * s[j + 1];
            if (wantv)
              for (int i = 0; i < n; i++) {
                t = cs * V[i][j] + sn * V[i][j + 1];
                V[i][j + 1] = -sn * V[i][j] + cs * V[i][j + 1];
                V[i][j] = t;
              }
            t = Maths.hypot(f, g);
            cs = f / t;
            sn = g / t;
            s[j] = t;
            f = cs * e[j] + sn * s[j + 1];
            s[j + 1] = -sn * e[j] + cs * s[j + 1];
            g = sn * e[j + 1];
            e[j + 1] = cs * e[j + 1];
            if (wantu && j < m - 1)
              for (int i = 0; i < m; i++) {
                t = cs * U[i][j] + sn * U[i][j + 1];
                U[i][j + 1] = -sn * U[i][j] + cs * U[i][j + 1];
                U[i][j] = t;
              }
          }
          e[p - 2] = f;
          iter = iter + 1;
        }
          break;

        // Convergence.

        case 4 : {

          // Make the singular values positive.

          if (s[k] <= 0.0) {
            s[k] = s[k] < 0.0 ? -s[k] : 0.0;
            if (wantv)
              for (int i = 0; i <= pp; i++)
                V[i][k] = -V[i][k];
          }

          // Order the singular values.

          while (k < pp) {
            if (s[k] >= s[k + 1])
              break;
            double t = s[k];
            s[k] = s[k + 1];
            s[k + 1] = t;
            if (wantv && k < n - 1)
              for (int i = 0; i < n; i++) {
                t = V[i][k + 1];
                V[i][k + 1] = V[i][k];
                V[i][k] = t;
              }
            if (wantu && k < m - 1)
              for (int i = 0; i < m; i++) {
                t = U[i][k + 1];
                U[i][k + 1] = U[i][k];
                U[i][k] = t;
              }
            k++;
          }
          iter = 0;
          p--;
        }
          break;
      }
    }
  }

  /*
   * ------------------------ Public Methods ------------------------
   */

  /**
   * Two norm condition number
   * 
   * @return max(S)/min(S)
   */

  public double cond() {
    return s[0] / s[Math.min(m, n) - 1];
  }

  /**
   * Return the diagonal matrix of singular values
   * 
   * @return S
   */

  public Matrix getS() {
    Matrix X = new Matrix(n, n);
    double[][] S = X.getArray();
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++)
        S[i][j] = 0.0;
      S[i][i] = s[i];
    }
    return X;
  }

  /**
   * Return the one-dimensional array of singular values
   * 
   * @return diagonal of S.
   */

  public double[] getSingularValues() {
    return s;
  }

  /**
   * Return the left singular vectors
   * 
   * @return U
   */

  public Matrix getU() {
    return new Matrix(U, m, Math.min(m + 1, n));
  }

  /**
   * Return the right singular vectors
   * 
   * @return V
   */

  public Matrix getV() {
    return new Matrix(V, n, n);
  }

  /**
   * Two norm
   * 
   * @return max(S)
   */

  public double norm2() {
    return s[0];
  }

  /**
   * Effective numerical matrix rank
   * 
   * @return Number of nonnegligible singular values.
   */

  public int rank() {
    double eps = Math.pow(2.0, -52.0);
    double tol = Math.max(m, n) * s[0] * eps;
    int r = 0;
    for (double element : s)
      if (element > tol)
        r++;
    return r;
  }
}
